Decomposition of Acceleration
(Page: 1
 2 ) Text We can approach curvature in terms of the circular images of the vectors T(t) and U(t). Since T(t) is a unit vector, it lies on the unit circle. So, there exists a function θ(t) such that T(t)=(cos(θ(t)), sin(θ(t))). Similarly, by the definition of normal vector, we have U(t) = (sin(θ(t)), cos(θ(t))). Then,
T'(t) = (sin(θ(t))θ'(t),cos(θ(t))θ'(t)) = U(t)θ'(t)
Entering this identity into the expression for acceleration yeilds X''(t) = s''(t)T(t) + s'(t)θ'(t)U(t). Earlier we defined the normal component of the accleration vector to be s'(t)2κg(t). Therefore, s'(t)θ'(t) = s'(t)2κg(t), or equivalently dθ/dt = (ds/dt)κg(t). Using an arclength parameterization, we then have
κg(s) = θ'(s)
This provides a definition of geodesic curvature in terms of the motion of the tangent and normal vectors on the unit circle.
Demos
Decomposition of Acceleration
 
This demonstration shows the decomposition of the acceleration vector at X(t0) into its tangential and normal components. It then graphs the geodesic curvature κg(t). In a third window, the unit tangent and and normal vectors are drawn in cyan and rad respectively. As you play the tapedeck for t0 in the control panel and the point X(t0) moves along the curve, notice how the speed of the tangent and normal vectors on the unit circle is related to the magnitude of the geodesic curvature.

